# Can a differential k-form be integrated on a manifold that is not k-dimensional?

For example, can you integrate a 2-form on some curve, a 1-dimensional manifold, or some 3-dimensional manifold?

I know that Stokes's Theorem states that if you integrate $\omega \in \mathcal A^{k-1}(M)$ or a (k-1)-differential form when integrated over the (k-1)-dimensional boundary of the k-dimensional manifold $M$, it is equal to integrating $d\omega \in \mathcal A^{k}(M)$ over the k-dimensional manifold $M$.

I.e. $\int_{\partial M} \omega = \int_{M} d\omega$.

I'm not very confident in all this because I'm new to learning university math, I am a novice in it and I'm doing this for fun as I am in Grade 11 still so I am not really forcing myself to learn all the details which can be bad. If you can recommend an article or book that explains my question that is suitable for my level, I would appreciate it a lot. I'm studying from Professor Shifrin's lectures on Multivariable Calculus. I was recently on one of his lectures on Stoke's Theorem and it was interesting because it seemed that all the dimensions of the forms and the manifolds (or its boundaries) matched up. Therefore, I was curious if I can integrate k-forms on manifolds of dimension less than k or bigger than k, as I can integrate at least some k-forms on k-dimensional manifolds according to Stokes's Theorem.

I have edited this question as I had no details whatsoever which can seem rude so I tried to put some context into it but I'm new to this site so please bear with me.

• No, you cannot.
– user147263
Commented Mar 20, 2015 at 22:53
• This is actually a good question. In certain situations, it does make sense to integrate a $k+\ell$-form over a $k$-dimensional manifold (or a collection of $k$-dimensional manifolds). It's called integration over the fiber. For example, consider the $2$-form $f(x,t)dt\wedge dx$ on $X\times [0,1]$. The result of integrating over the fiber (in this case, $[0,1]$) is the $1$-form $\big(\int_0^1 f(x,t)dt\big)dx$ on $X$. This has all sorts of topological interpretations, too. Commented Mar 21, 2015 at 2:08
• @TedShifrin: Thank you, I appreciate your comment and I want to thank you for your videos that helped me learn some multivariable calculus. I am by no means an expert and I have edited the question to fill in the context of my question. However, I am not knowledgeable enough to get your comment. Can you just confirm that for special cases, you can integrate over manifolds with different dimension than the forms? Are you saying that $l$ can be positive or negative? Thank you. Commented Mar 21, 2015 at 2:42
• Ordinarily dimensions must match up. But in certain situations which you may see when you're far more advanced, you can integrate a $k+\ell$ form over $k$-dimensional "stripes" and be left with an $\ell$-form on the stuff perpendicular to them. Get back to me in a few years :) For now, make the dimensions match! :) By the way, Stokes's Theorem, while super important, is sort of a red herring, so far as your question is concerned. Commented Mar 21, 2015 at 2:52
• No, there's no notion, unless you give me lots of extra structure (like a particularly nice foliation by $p$-dimensional manifolds, with the leaf space itself being a manifold. This reduces to the case I already discussed. Commented Feb 2, 2019 at 0:28

A simple way in which this may be realized: By taking the wedge product of the $k$-form $\omega$ with an $l$-form $\alpha$ one arrives at a $(k+l)$-form $\alpha\wedge\omega$ which may be integrated over a $(k+l)$-dimensional space. By taking the interior product of $\omega$ with $l$ vector fields one arrives at a $(k-l)$-form which may be integrated over a $(k-l)$-dimensional space.